I mainly see this in physics, such as the proofs to a certain formula. the question is asking for $$\left(x-\frac{y}{2}\right)^{-3} - \left(x + \frac{y}{2}\right)^{-3}$$,
but I used common denominators to reach $$\left(x + \frac{y}{2}\right)^3 - \left(x - \frac{y}{2}\right)^3$$ on the numerator. Anyways, the result should be $$3x^2y + \frac{y^3}{4}$$ as per usual, but given the condition $x \gg y$, the textbook seemingly ignores the $\frac{y^3 }{ 4}$ term. This is my main source of confusion as to when a certain term can be neglected, and I'd appreciate any explanation. If we were to approach it using negative binomial expansion, is the common practice to stop two terms in? (i.e $$\left(x - \frac{y}{2}\right)^{-3} = x^{-3} - 3x^{-4}\left(\frac{-y}{2} \right)$$
If I properly read the question (please, start to use MathJax), you have $$z=\frac{1}{\left(x+\frac{y}{2}\right)^3}-\frac{1}{\left(x-\frac{y}{ 2}\right)^3}$$ Let $y=kx$ and simplify to make $$z=\frac 1 {x^3} \left( \frac{1}{\left(1+\frac{k}{2}\right)^3}-\frac{1}{\left(1-\frac{k}{2}\right)^3}\right)$$ Use binomial expansion or Taylor series or long division $$(1+\epsilon)^{-3}=1-3 \epsilon +6 \epsilon ^2-10 \epsilon ^3+O\left(\epsilon ^4\right)$$
Replace $\epsilon$ by $\pm \frac k2$ to obtain $$z=\frac 1 {x^3} \left( -3 k-\frac{5 }{2}k^3+O\left(k^4\right) \right)$$ SInce $x\gg y$, $k\ll 1$ and $k^3$ is negigible.
So at first order $$z\sim -3\frac k{x^3}= -\frac 3{x^4}$$
Is this clear for you ? If not, tell me.